Year: 2009
Paper: 3
Question Number: 4
Course: LFM Pure
Section: Integration
The vast majority of candidates (in excess of 95%) attempted at least five questions, and nearly a quarter attempted more than six questions, though very few doing so achieved high scores (about 2%). Most attempting more than six questions were submitting fragmentary answers, which, as the rubric informed candidates, earned little credit.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
For any given (suitable) function $\f$, the \textit{Laplace transform} of $\f$ is the function $\F$ defined by
\[
\F(s) = \int_0^\infty \e^{-st}\f(t)\d t
\quad \quad \, (s>0)
\,.
\]
\begin{questionparts}
\item
Show that the Laplace transform of $\e^{-bt}\f(t)$, where $b>0$,
is $\F(s+b)$.
\item Show that
the Laplace transform of $\f(at)$, where $a>0$, is $a^{-1}\F(\frac s a)\,$.
\item Show that the Laplace
transform of $\f'(t)$ is $s\F(s) -\f(0)\,$.
\item
In the case $\f(t)=\sin t$, show that $\F(s)= \dfrac 1 {s^2+1}\,$.
\end{questionparts}
Using only these four results, find the Laplace transform of
$\e^{-pt}\cos{qt}\,$, where $p>0$ and $q>0$.\begin{questionparts}
\item \begin{align*}
\mathcal{L}\{e^{-bt}f(t)\}(s) &= \int_0^{\infty}e^{-st}\{ e^{-bt}f(t) \} \d t \\
&= \int_0^{\infty} e^{-(s+b)t}f(t) \d t \\
&= F(s+b)
\end{align*}
\item \begin{align*}
\mathcal{L}\{f(at)\}(s) &= \int_0^{\infty} e^{-st}f(at) \d t \\
&= \int_{u=0}^{\infty}e^{-s \frac{u}{a}} f\left(a \tfrac{u}{a}\right)\frac{1}{a} \d u \\
&= \int_0^{\infty}e^{-su/a}f(u) a^{-1} \d u \\
&= a^{-1} \int_0^{\infty} e^{-(s/a)u}f(u) \d u \\
&= a^{-1} F\left (\frac{s}{a} \right)
\end{align*}
\item \begin{align*}
\mathcal{L}\{f'(t)\}(s) &= \int_0^{\infty} e^{-st}f'(t) \d t \\
&= \left [e^{-st} f(t) \right]_0^{\infty} - \int_0^{\infty} -s e^{-st} f(t) \d t\\
&= -f(0)+sF(s) \\
&= sF(s) - f(0)
\end{align*}
\item Since $f''(t) = -f(t)$ we must have:
\begin{align*}
&& -\mathcal{L}(f)&= \mathcal{L}(f'') \\
&&&= s\mathcal{L}(f') -f'(0) \\
&&&= s(s\mathcal{L}(f)-f(0)) - f'(0) \\
&&&= s^2\mathcal{L}(f) - 1 \\
\Rightarrow && (1+s^2) \mathcal{L}(f) &= 1 \\
\Rightarrow && F(s) &= \frac{1}{1+s^2}
\end{align*}
\end{questionparts}
\begin{align*}
\mathcal{L}\{e^{-pt}\cos qt\}(s) &= \mathcal{L}\{\cos qt\}(s+p) \\
&= q^{-1}\mathcal{L}\{\cos t\}\left (\frac{s+p}{q} \right) \\
&= q^{-1}\mathcal{L}\{\sin'\}\left (\frac{s+p}{q} \right) \\
&= q^{-1} \left (\frac{s+p}{q} \right) \mathcal{L}\{\sin\} \left (\frac{s+p}{q} \right) - q^{-1}\sin \left (0\right) \\
&= \frac{s+p}{q^2} \frac{1}{1+\left (\frac{s+p}{q} \right)^2 } \\
&= \frac{s+p}{q^2+(s+p)^2}
\end{align*}
About half the candidates attempted this, with similar levels of success to question 3. Parts (i) and (iii) caused few problems though part (ii) did. There were some errors in part (iv), but it was the last part using the four results that usually went wrong.